IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

barsharahul

_{[X ]}

_[

_}(2^x/ X²) =??

in words the question is "what is the value of Limit (2^x/ X²) where x tends to infinity".

elessar_iitkgp

_x

(2^x/x²) (Of the form

/

. Hence we apply the L'Hospital's Rule)
=_x

(2^x ln2/2x) (Again

/

. Again applying LH Rule)
_=x

(2^x(ln2)²/2)
=

priyesh

can you please explain what is LH rule as i'm unaware of it

iitkgp_bipin

L'Hopitals Rule says :

If the limit _[x]

_[c] f(x)/g(x) is in the form of 0/0 or

/

, then we can apply

_[x]

_[c] f(x)/g(x) = _[x]

_[c] f'(x)/g'(x)

L = _[x]

_[infinity] 2^x/x² which is in

/

form.

Applying LH rule :
L = _[x]

_[infinity] 2^xln2/2x which is again in

/

form.

Again applying LH rule :
L = _[x]

_[infinity] 2^x(ln2)²/2 which tends to

.